B. Let $1\le p \le \infty$. Fix $p$. Let $f:D\to C$ be a holomorphic function. Then must there exists a point $q\in S^1$ such that in some neighborhood $U $ of $q$ in $D$, $f \in L^p(U)$.i.e., on the contrary, is it possible to have a holomorphic function $f$ not belonging to $ L^p(U) \forall q \in S^1$ and $\forall $ neighborhood $U$ of $q$ ?

If I were you, I would study the equivalent (locally) problem of boundary behavior of harmonic functions on the upper half plane. Then you have a very convenient representation of the functions via the Poisson kernel; you can prescribe an arbitrary trace (e.g. a positive function belonging only to some Lp spaces but not others) and you get a positive harmonic function which converges to the trace monotonically
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Piero D'AnconaAug 14 '11 at 9:11

3 Answers
3

For the first question the answer is positive. Consider the function $f(x)=\sum_{n=1}^\infty x^{2^n}$. Then $f(x^2)=f(x)-x$. Since $\lim_{x\to1-0}f(x)=+\infty$ then the same is true as $x=re^{i \pi\psi }\to q=e^{i \pi\psi }$ where $\psi$ is any binary rational point (of the form $k/2^n$, $k\in \mathbb Z$, $n\in \mathbb Z_+$). Approximating now an arbitrary $\varphi$ by such numbers we will have the statement for any $q=e^{i\pi\varphi}$.

Suppose f is in $L^p(U)$. Then $f$ has a trace on the boundary that is in $W^{-1/p,p}$.
You can conclude this, for instance, from the trace theorem for divergence free vector fields: If f=u+iv, then the vector fields (u,-v) and (v,u) are both divergence free, so their normal components have a trace on the boundary.

On the other hand, you can prescribe a distribution on the boundary that is not in $W^{-1/p,p}$ on any subinterval of the boundary, and then find a harmonic function which takes on these boundary values.